View Full Version : A box with no top. Minimize cardboard used. How???

Kristy

07-21-2007, 08:38 PM

#10 Directions:

A box with a square base and open top must have a volume of 32,000 cm^3.

Fnd the dimensions of the box that minimize the amount of material used.

Ok it would have a bottom that would be Length x Width,

2 sides that would be Length x Height

2 sides that would be Width x Height

Volume = L*W*H =32,000 cm 3

Surface area (without top) = L*W + 2 L * H + 2 W x H = f (something)

Then I know I’m supposed to take the derivative, but how do I do that with three variables.

galactus

07-21-2007, 09:03 PM

Since the base is square you only have two variables.

The volume is \L\\V=x^{2}y=32,000.........[1]

The surface area is \L\\S=x^{2}+4xy.........[2]

You are minimizing [2]. So, solve [1] for y and sub into [2]. Then do the differentiating, set to 0 thing and solve for x.

Kristy

07-21-2007, 09:07 PM

#10 Directions:

A box with a square base and open top must have a volume of 32,000 cm^3.

Fnd the dimensions of the box that minimize the amount of material used.

Thanks. I kinda missed the "square base" part too. That would have made it more obvious. Oops. As soon as I saw you x^2 I wondered why you could do that, and then....okay its square. :(

Okay, makes much more sense now.

Kristy

07-21-2007, 09:14 PM

Since the base is square you only have two variables.

The volume is \L\\V=x^{2}y=32,000.........[1]

The surface area is \L\\S=x^{2}+4xy.........[2]

You are minimizing [2]. So, solve [1] for y and sub into [2]. Then do the differentiating, set to 0 thing and solve for x.

solve [1] for y

The volume is \L\\V=x^{2}y=32,000.........[1]

\L\\y = \frac{32,000}{x^{2}}

and sub into [2]

The surface area is \L\\S=x^{2}+4xy.........[2]

The surface area is \L\\S=x^{2}+4x(\frac{32,000}{x^{2}}).........[2]

The surface area is \L\\S=x^{2}+4(\frac{32,000}{x}).........[2]

The surface area is \L\\S=x^{2}+(\frac{128,000}{x}).........[2]

The surface area is \L\\S=x^{2}+128,000(\frac{1}{x}).........[2]

in progress...wow this tex stuff is hard! Almost as haard as the math!

galactus

07-21-2007, 09:15 PM

Yep. Now sub that into the surface area equation. It'll be entirely in terms of x and you can differentiate and all that.

Kristy

07-21-2007, 09:25 PM

Sorry I kept editing my post because of trying to make sure I was doing the tex right.

Since the base is square you only have two variables.

The volume is \L\\V=x^{2}y=32,000.........[1]

The surface area is \L\\S=x^{2}+4xy.........[2]

You are minimizing [2]. So, solve [1] for y and sub into [2]. Then do the differentiating, set to 0 thing and solve for x.

solve [1] for y

The volume is \L\\V=x^{2}y=32,000.........[1]

\L\\y = \frac{32,000}{x^{2}}

and sub into [2]

The surface area is \L\\S=x^{2}+4xy.........[2]

The surface area is \L\\S=x^{2}+4x(\frac{32,000}{x^{2}}).........[2]

The surface area is \L\\S=x^{2}+4(\frac{32,000}{x}).........[2]

The surface area is \L\\S=x^{2}+(\frac{128,000}{x}).........[2]

The surface area is \L\\S=x^{2}+128,000(\frac{1}{x}).........[2]

in progress...wow this tex stuff is hard! Almost as haard as the math!

The surface area is \L\\S=x^{2}+128,000(x^{-1}).........[2]

Then do the differentiating, set to 0 thing and solve for x

S'(x) = 2x + 128,000 * (-1)x^{-2}

S'(x) = 2x + 128,000 *\frac{-1}{x^{2}}

0 = 2x + 128,000 *\frac{-1}{x^{2}}

\frac{128,000}{x^{2}} = \frac{2x}{1}

2x^{3}=128,000

divide both sides by 2

x^{3}=64,000

take the cube root of both sides

(is there a negative cube root too or is that just with squre roots?

x=4,000

So that means the bottom is 4,000X4,000

the height is y and is

\L\\y = \frac{32,000}{x^{2}}

=\L\\y = \frac{32,000}{4^{2}}

y = 2000

So the dimensions are 4000 length and with and 2000 height.

Does that seem right?

galactus

07-21-2007, 09:40 PM

You're doing good until S'(x)=2x.

What about differentiating \L\\\frac{128000}{x}?.

After you have it, set to 0 and solve for x.

Kristy

07-21-2007, 10:25 PM

Thanks for the encouragement. I think I got it now, not sure though. I guess a box could be twice as long and wide and not very tall, I'm not really sure how to check this one though.

galactus

07-21-2007, 11:27 PM

You get:

\L\\S'(x)=2x-\frac{128000}{x^{2}}

Set to 0 and solve for x:

\L\\2x-\frac{128000}{x^{2}}=0

\L\\2x^{3}=128000

x=40, \;\ and \;\ y=20

See?. That wasn't hard. :D It that what you got?.

Kristy

07-22-2007, 12:09 AM

You get:

\L\\S'(x)=2x-\frac{128000}{x^{2}}

Set to 0 and solve for x:

\L\\2x-\frac{128000}{x^{2}}=0

\L\\2x^{3}=128000

x=40, \;\ and \;\ y=20

See?. That wasn't hard. :D It that what you got?.

I got extra 0's for some reason I don't know.

I got x to be 4000 and y 2000 see my post above.

I do'nt know what I did to multiply everythhing by 100

galactus

07-22-2007, 09:40 AM

The cube root of 64000 isn't 4000, it's 40. That's why you're off by a multiple of 100.

Kristy

07-22-2007, 03:40 PM

The cube root of 64000 isn't 4000, it's 40. That's why you're off by a multiple of 100.

Oops. What I did was take the cube root of 64, found it to be 4, and then added the 000 so it was 4000.

So the dimensions are x = 40 length and width

height y = 20

?

Subhotosh Khan

08-04-2007, 12:26 PM

(is there a negative cube root too or is that just with squre roots?

Positive numbers can have only positive cube-roots.

Negative numbers can have only negative cube-roots.

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